What is the derivative of (x^3+3x)/(x^2-4x+3) ?

The derivative of (x^3+3x)/(x^2-4x+3) is (x^4-8x^3+6x^2+9)/((x^2-4x+3)^2)

What is the first derivative of (x^3+3x)/(x^2-4x+3) ?

The first derivative of (x^3+3x)/(x^2-4x+3) is (x^4-8x^3+6x^2+9)/((x^2-4x+3)^2)

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derivative of

\frac{x ^{3} + 3 x}{x ^{2} - 4 x + 3}

\frac{x ^{4} - 8 x ^{3} + 6 x ^{2} + 9}{( x ^{2} - 4 x + 3 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{3} + 3 x}{x ^{2} - 4 x + 3})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( x ^{3} + 3 x ) ( x ^{2} - 4 x + 3 ) - \frac{d}{d x} ( x ^{2} - 4 x + 3 ) ( x ^{3} + 3 x )}{( x ^{2} - 4 x + 3 ) ^{2}}

\frac{d}{d x} (x^{3} + 3 x) = 3 x^{2} + 3

\frac{d}{d x} (x^{2} - 4 x + 3) = 2 x - 4

= \frac{( 3 x ^{2} + 3 ) ( x ^{2} - 4 x + 3 ) - ( 2 x - 4 ) ( x ^{3} + 3 x )}{( x ^{2} - 4 x + 3 ) ^{2}}

Expand

(3 x^{2} + 3) (x^{2} - 4 x + 3) - (2 x - 4) (x^{3} + 3 x) : x^{4} - 8 x^{3} + 6 x^{2} + 9

= \frac{x ^{4} - 8 x ^{3} + 6 x ^{2} + 9}{( x ^{2} - 4 x + 3 ) ^{2}}

x \to 4 - lim (\frac{∣ x - 4 ∣}{x - 4})

\int (x^{3} + x) x^{4} + 2 x^{2} d x

\frac{\partial}{\partial x} (sin (x z^{2}) - x y^{2})

\int x a e^{- a x} d x

y^{^{'}} + 3 t y = t, y (0) = 1

What is the derivative of (x^3+3x)/(x^2-4x+3) ?
The derivative of (x^3+3x)/(x^2-4x+3) is (x^4-8x^3+6x^2+9)/((x^2-4x+3)^2)
What is the first derivative of (x^3+3x)/(x^2-4x+3) ?
The first derivative of (x^3+3x)/(x^2-4x+3) is (x^4-8x^3+6x^2+9)/((x^2-4x+3)^2)